迭代向量的笛卡尔积 [英] Iterate over cartesian product of vectors

问题描述

我有以下嵌套循环:

for (x in xs) {
    for (y in ys) {
        # Do something with x and y
    }
}

我想要展平，所以我想建立两个向量 xs 和 ys 的笛卡尔积并迭代结果.在 Python 中，这将是微不足道的:

Which I’d like to flatten so I thought of building a Cartesian product of the two vectors xs and ys and iterating over the result. In Python, this would be trivial:

for xy in product(xs, ys):
    # x, y = xy[0], xy[1]

但在 R 中，我发现的最简单的等价物看起来令人生畏:

But in R, the simplest equivalent I’ve found looks daunting:

xys <- expand.grid(xs, ys)
for (i in 1 : nrow(xys)) {
    xy <- as.vector(xys[i, ])
    # x <- xy[1], y <- xy[2]
}

肯定有更好的方法，不是吗?(澄清一下，我不想迭代索引……我认为必须有一种方法可以直接迭代产品中的元组.)

Surely there must be a better way, no? (To clarify, I don’t want to iterate over an index … I think there must be a way to directly iterate over the tuples in the product.)

推荐答案

您可以使用 apply 函数将函数应用于数据框的每一行.只需将 "your function" 替换为您的实际功能即可.

You can use the apply function to apply a function to each row of your data frame. Just replace "your function" with your actual function.

# example data
xs <- rnorm(10)
ys <- rnorm(10)    

apply(expand.grid(xs, ys), 1, FUN = function(x) {"your function"})

这是一个非常基本的例子.这里，计算了一行中两个值的总和:

This is a very basic example. Here, the sum of both values in a row is calculated:

apply(expand.grid(xs, ys), 1, FUN = function(x) {x[1] + x[2]})

<小时>

这是一个使用命名参数 (xs, ys) 而不是索引 (x[1], x[2]):

---

Here is a variant that uses named arguments (xs, ys) instead of indices (x[1], x[2]):

myfun <- function(xs, ys) xs + ys
arguments <- expand.grid(xs = rnorm(10), ys = rnorm(10))
apply(arguments, 1, function(x)do.call(myfun, as.list(x)))

这篇关于迭代向量的笛卡尔积的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！